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I cannot find a simple explanation of the damping ratio formula.

The natural frequency for a spring mass system seems pretty simple:

position, velocity and acceleration are given by: $$x(t)=Acos(\omega t)$$ $$v=x'(t)=-A\omega sin(\omega t)$$ $$a=x''(t)=-A\omega^2 cos(\omega t)$$

Replacing a and x in $ma = -kx$ with the formulas above, we have: $$-mA\omega^2 cos(\omega t)=-kAcos(\omega t)$$

And therefore $m\omega^2=k$, from which we get $m\omega^2=k$ and hence:

$$\omega=\sqrt{\frac{k}{m}}$$

How do we get to the following formula for the damping ratio?

$$\frac{b}{m}=2\zeta\omega$$

Pinco Pallino
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1 Answers1

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We have $2 \zeta \omega= \frac{b}{m}$ by definition, $\zeta$ is just defined to be this quantity.

M10687
  • 3,250
  • That wasn't my question... the question was more how do we get to $\frac{b}{m}=2\zeta \omega$ at all. Some reasoning like the one that led to $m \omega^2 = k$. – Pinco Pallino Jun 05 '16 at 20:52
  • I edited the answer, $\zeta$ is just defined this way, there no way to "derive" that equation. It may help to read: https://en.wikipedia.org/wiki/Damping_ratio – M10687 Jun 05 '16 at 21:00